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2 years ago

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Two old RSM students were standing ten feet apart when they ran in opposite directions with speeds of 180 feet per minute and 16

0 feet per minute respectively, how many minutes would pass before they were 1870 feet apart?

1 answer:

inessss [21]2 years ago

7 0

The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes

Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.

Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'

D = 180t + 10 + 160t

D = 340t + 10

<h3>Number of minutes before they are 1870 feet</h3>

Making t subject of the formula, we have

t = (D - 10)/340

Since they are 1870 feet apart after t minutes, D = 1870 feet.

t = (D - 10)/340

t = (1870 - 10)/340

t = 1860/340

t = 5.47 minutes

So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes

Learn more about minutes of distance apart here

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Given:

The data set is

10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 35

To find:

Effect on mean, median, range after removing the outlier, 35.

Solution:

We have, the data set

10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17, 35

$\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}$

$\text{Mean}=\dfrac{10+10+11+12+13+13+13+14+14+15+15+15+16+17+17+17+35}{17}$

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$Median=\dfrac{n+1}{2}\text{th term}$ because n=17, which is odd.

$Median=\dfrac{17+1}{2}\text{th term}$

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After removing the outlier, 35, the data set is

10, 10, 11, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 17, 17, 17

$\text{Mean}=\dfrac{10+10+11+12+13+13+13+14+14+15+15+15+16+17+17+17}{16}$

$\text{Mean}=\dfrac{222}{16}$

$\text{Mean}=13.875$

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$Median=\dfrac{\dfrac{n}{2}\text{th term}+(\dfrac{n}{2}+1)\text{th term}}{2}$ because n=16, which is even.

$Median=\dfrac{\dfrac{16}{2}\text{th term}+(\dfrac{16}{2}+1)\text{th term}}{2}$

$Median=\dfrac{8\text{th term}+9\text{th term}}{2}$

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$Median=\dfrac{28}{2}$

$Median=14$

New median is 14, which is same as original median.

$Range=17-10$

$Range=7$

So, the new range is 7 which is less than 25.

Therefore, the mean of the data set will decrease, the median of the data set will not change, the Range of the data set will decrease.

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